Charles K. Alexander, Matthew N.O. Sadiku, Circuiti elettrici, 4e
© 2014 McGraw-Hill Education (Italy) srl
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(VHUFL]LDULR
Capitolo 5
5.1
(a)
(b)
(c)
Rin = 1.5 MΩ
Rout = 60 Ω
A = 8x104
Therefore AdB = 20 log 8x104 = 98.0 dB
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5.2
v0 = Avd = A(v2 - v1)
= 105 (20-10) x 10-6 = 1V
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.3
v0 = Avd = A(v2 - v1)
= 2 x 105 (30 + 20) x 10-6 = 10V
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.4
v0 = Avd = A(v2 - v1)
v
−4
= −2µV
v2 - v 1 = 0 =
A 2 x10 6
v2 - v1 = -2 µV = –0.002 mV
1 mV - v1 = -0.002 mV
v1 = 1.002 mV
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.5
I
R0
Rin
vd
+
vi
Avd
+
+
v0
-
-
-vi + Avd + (Ri + R0) I = 0
But
+
-
(1)
vd = RiI,
-vi + (Ri + R0 + RiA) I = 0
I=
vi
R 0 + (1 + A )R i
(2)
-Avd - R0I + v0 = 0
v0 = Avd + R0I = (R0 + RiA)I =
(R 0 + R i A) v i
R 0 + (1 + A)R i
v0
R 0 + RiA
100 + 10 4 x10 5
=
=
⋅ 10 4
5
v i R 0 + (1 + A)R i 100 + (1 + 10 )
≅
10 9
100,000
⋅ 10 4 =
= 0.9999990
5
100,001
1 + 10
(
)
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.6
vi
+ -
R0
I
Rin
vd
+
-
+
Avd
+
vo
-
(R0 + Ri)R + vi + Avd = 0
But
vd = RiI,
vi + (R0 + Ri + RiA)I = 0
I=
− vi
R 0 + (1 + A)R i
(1)
-Avd - R0I + vo = 0
vo = Avd + R0I = (R0 + RiA)I
Substituting for I in (1),
 R 0 + RiA 
 vi
v0 = − 
 R 0 + (1 + A)R i 
50 + 2x10 6 x 2 x10 5 ⋅ 10 −3
= −
50 + 1 + 2x10 5 x 2 x10 6
(
≅
(
)
)
− 200,000 x 2 x10
mV
200,001x 2 x10 6
6
v0 = -0.999995 mV
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.7
100 kΩ
10 kΩ
VS
+
–
Rout = 100 Ω
1
2
+
Vd
Rin
–
+
AVd
–
At node 1,
+
Vout
–
(VS – V1)/10 k = [V1/100 k] + [(V1 – V0)/100 k]
10 VS – 10 V1 = V1 + V1 – V0
which leads to V1 = (10VS + V0)/12
At node 2,
(V1 – V0)/100 k = (V0 – (–AVd))/100
But Vd = V1 and A = 100,000,
V1 – V0 = 1000 (V0 + 100,000V1)
0= 1001V0 + 99,999,999[(10VS + V0)/12]
0 = 83,333,332.5 VS + 8,334,334.25 V0
which gives us (V0/ VS) = –10 (for all practical purposes)
If VS = 1 mV, then V0 = –10 mV
Since V0 = A Vd = 100,000 Vd, then Vd = (V0/105) V = –100 nV
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.8
(a)
If va and vb are the voltages at the inverting and noninverting terminals of the op
amp.
va = vb = 0
1mA =
0 − v0
2k
v0 = -2V
(b)
10 kΩ
2V
+
ia
va
2V
+
vb
1V
+
vo
+
2 kΩ
-
-
+
va
-
10 kΩ
+-
+
ia
vo
-
(b)
(a)
Since va = vb = 1V and ia = 0, no current flows through the 10 kΩ resistor. From Fig. (b),
-va + 2 + v0 = 0
v0 = va - 2 = 1 - 2 = -1V
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.9
(a)
Let va and vb be respectively the voltages at the inverting and noninverting
terminals of the op amp
va = vb = 4V
At the inverting terminal,
1mA =
4 − v0
2k
v0 = 2V
(b)
1V
+-
+
+
vb
vo
-
-
Since va = vb = 3V,
-vb + 1 + vo = 0
vo = vb - 1 = 2V
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.10
Since no current enters the op amp, the voltage at the input of the op amp is vs. Hence
 10  v o
vs = vo 
=
 10 + 10  2
vo
=2
vs
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.11
8 kΩ
2 kΩ
3V
vb =
+
−
5 kΩ
a
b
−
+
10 kΩ
io
+
4 kΩ
vo
−
10
(3) = 2V
10 + 5
At node a,
3 − va va − vo
=
2
8
12 = 5va – vo
But va = vb = 2V,
12 = 10 – vo
–io =
vo = –2V
va − vo 0 − vo 2 + 2 2
+
=
+ = 1mA
8
4
8
4
i o = –1mA
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.12
This is an inverting amplifier.
25
vo = −
vs

→
5
vo
= −5
vs
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.13
10 kΩ
a
b
1V
+
−
+
−
io
100 kΩ i2
90 kΩ
i1
+
10 kΩ
vo
50 kΩ
By voltage division,
va =
90
(1) = 0.9V
100
vb =
v
50
vo = o
150
3
But va = vb
io = i1 + i2 =
v0
= 0.9
3
vo = 2.7V
vo
v
+ o = 0.27mA + 0.018mA = 288 µA
10k 150k
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
−
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5.14
Transform the current source as shown below. At node 1,
10 − v1 v1 − v 2 v1 − v o
=
+
5
20
10
10 kΩ
5 kΩ
10 kΩ
20 kΩ
v1
10V
vo
v2
+
−
−
+
+
vo
−
But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo
At node 2,
v1 − v 2 v 2 − v o
=
,
20
10
40 = 7v1 - 2vo
v 2 = 0 or v1 = -2vo
From (1) and (2), 40 = -14vo - 2vo
(1)
(2)
vo = -2.5V
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.15
(a) Let v1 be the voltage at the node where the three resistors meet. Applying
KCL at this node gives
 1
v1 v1 − vo
1  vo
 −
+
= v1 
+
R2
R3
 R2 R3  R3
At the inverting terminal,
is =
0 − v1

→ v1 = −i s R1
R1
Combining (1) and (2) leads to

v
R
R 

→
i s 1 + 1 + 1  = − o
R3
 R2 R3 
is =
(1)
(2)

RR 
vo
= − R1 + R3 + 1 3 
R2 
is

(b) For this case,
vo
20 x 40 

= − 20 + 40 +
 kΩ = - 92 kΩ
is
25 

C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.16
10k Ω
5k Ω
ix
va
vb
+
0.5V
-
iy
+
vo
2k Ω
8k Ω
Let currents be in mA and resistances be in k Ω . At node a,
0.5 − v a v a − vo
=

→ 1 = 3v a − vo
5
10
(1)
But
8
10
(2)
vo

→ vo = v a
8+2
8
Substituting (2) into (1) gives
10
8
1 = 3v a − v a

→ v a =
8
14
Thus,
0.5 − v a
ix =
= −1 / 70 mA = − 14.28 µA
5
v − vb v o − v a
0.6 8
10
iy = o
+
= 0.6( v o − v a ) = 0.6( v a − v a ) =
x mA = 85.71 µA
4 14
8
2
10
v a = vb =
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.17
(a)
(b)
(c)
G=
vo
R
12
= − 2 = − = -2.4
vi
R1
5
vo
80
=−
= -16
vi
5
vo
2000
=−
= -400
vi
5
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.18
We temporarily remove the 20-kΩ resistor. To find VTh, we consider the circuit below.
10 kΩ
2 kΩ
12 kΩ
–
+
2 mV
+
+
_
8Ω
VTh
–
This is an inverting amplifier.
10k
VTh = −
(2mV ) = −10mV
2k
To find RTh, we note that the 8-kΩ resistor is across the output of the op amp which is
acting like a voltage source so the only resistance seen looking in is the 12-kΩ resistor.
The Thevenin equivalent with the 20-kΩ resistor is shown below.
12 kΩ
a
I
–10 mV
+
_
20 k
b
I = –10m/(12k + 20k) = 0.3125x10–6 A
p = I2R = (0.3125x10–6)2x20x103 = 1.9531 nW
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.19
We convert the current source and back to a voltage source.
24=
(4/3) kΩ
(2/3)V
4 kΩ
0V
4
3
10 kΩ
+
−
−
+
vo
5 kΩ
10k  2 
  = -1.25V
4  3

 4 + k
3

v
v −0
io = o + o
= -0.375mA
5k
10k
vo = −
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.20
8 kΩ
4 kΩ
9V
a
+
−
4 kΩ
vs
2 kΩ
b
−
+
+
−
+
vo
−
At node a,
9 − va va − vo va − vb
=
+
4
8
4
18 = 5va – vo - 2vb
(1)
At node b,
va − vb vb − vo
=
4
2
va = 3vb - 2vo
(2)
But vb = vs = 0; (2) becomes va = –2vo and (1) becomes
-18 = -10vo – vo
vo = -18/(11) = -1.6364V
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.21
Let the voltage at the input of the op amp be va.
va = 1 V,
3-v a va − vo
=
4k
10k

→
3-1 1− vo
=
4
10
vo = –4 V.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.22
Av = -Rf/Ri = -15.
If Ri = 10kΩ, then Rf = 150 kΩ.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.23
At the inverting terminal, v=0 so that KCL gives
vs − 0
0 0 − vo
=
+
R1
R2
Rf
→
vo
vs
=−
Rf
R1
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.24
v1
Rf
R1
R2
- vs +
+
+
R4
vo
-
R3
v2
We notice that v1 = v2. Applying KCL at node 1 gives
v1 (v1 − v s ) v1 − vo
+
+
=0
R1
R2
Rf

→
 1

 + 1 + 1 v1 − v s = vo
R R
R2 R f
R f 
2
 1
(1)
Applying KCL at node 2 gives
R3
v1 v1 − v s
+
=0

→ v1 =
vs
R3
R4
R3 + R4
Substituting (2) into (1) yields
(2)
 R
R
R  R3  1 
 −  v s
vo = R f  3 + 3 − 4 


R
R
R
R
+
R
R2 
f
2  3
4 
 1
i.e.
 R
R
R  R3  1 
 − 
k = R f  3 + 3 − 4 


+
R
R
R
R
R
R2 
 1
f
2  3
4 
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.25
This is a voltage follower. If v1 is the output of the op amp,
v1 = 2V
vo =
20k
20
v1= (12)=1.25 V
20k+12k
32
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.26
+
vb
+
0.4V
-
-
io
+
8k Ω
2k Ω
5k Ω
vo
-
vb = 0.4 =
8
vo = 0.8vo
8+ 2

→
vo = 0.4 / 0.8 = 0.5 V
Hence,
io =
v o 0.5
=
= 0.1 mA
5k 5k
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.27
This is a voltage follower.
24
(5) = 3V , v2 = v1 = 3V
24 + 16
12
vo =
(3V ) = 1.8 V
12 + 8
v1 =
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.28
50 kΩ
v1
−
+
va
10 kΩ
At node 1,
+
−
0.4 V
vo
20 kΩ
0 − v1 v1 − v o
=
10k
50k
But v1 = 0.4V,
-5v1 = v1 – vo, leads to
vo = 6v1 = 2.4V
Alternatively, viewed as a noninverting amplifier,
vo = (1 + (50/10)) (0.4V) = 2.4V
io = vo/(20k) = 2.4/(20k) = 120 µA
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.29
R1
va
vb
+
vi
-
+
-
R2
+
R2
vo
R1
-
va =
R2
vi ,
R1 + R2
But v a = vb
vb =

→
R1
vo
R1 + R2
R2
R1
vi =
vo
R1 + R2
R1 + R2
Or
v o R2
=
vi
R1
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.30
The output of the voltage becomes
vo = vi = 12
30 20 = 12kΩ
By voltage division,
vx =
12
(1.2) = 0.2V
12 + 60
ix =
vx
0.2
=
= 10µA
20k 20k
p=
v 2x 0.04
=
= 2µW
R
20k
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.31
After converting the current source to a voltage source, the circuit is as shown below:
12 kΩ
3 kΩ
1
6 kΩ v
o
v1
12 V
+
−
+
−
2
vo
6 kΩ
At node 1,
12 − v1 v1 − v o v1 − v o
=
+
3
6
12
48 = 7v1 - 3vo
(1)
At node 2,
v1 − v o v o − 0
=
= ix
6
6
v1 = 2vo
(2)
From (1) and (2),
48
11
v
i x = o = 727.2µA
6k
vo =
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
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5.32
Let vx = the voltage at the output of the op amp. The given circuit is a non-inverting
amplifier.
 50 
v x = 1 +  (4 mV) = 24 mV
 10 
60 30 = 20kΩ
By voltage division,
v
20
v x = x = 12mV
20 + 20
2
vx
24mV
ix =
=
= 600nA
(20 + 20)k 40k
vo =
p=
v o2 144x10 −6
=
= 204nW
R
60x10 3
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.33
After transforming the current source, the current is as shown below:
1 kΩ
4 kΩ
4V
vi
+
−
va
+
−
2 kΩ
vo
3 kΩ
This is a noninverting amplifier.
3
 1
v o = 1 +  v i = v i
2
 2
Since the current entering the op amp is 0, the source resistor has a OV potential drop.
Hence vi = 4V.
vo =
3
(4) = 6V
2
Power dissipated by the 3kΩ resistor is
v o2 36
=
= 12mW
R 3k
ix =
va − vo 4 − 6
=
= -2mA
R
1k
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.34
v1 − vin v1 − vin
+
=0
R1
R2
(1)
R3
vo
R3 + R 4
(2)
but
va =
Combining (1) and (2),
v1 − va +
R1
R
v 2 − 1 va = 0
R2
R2

R 
R
v a 1 + 1  = v1 + 1 v 2
R2
 R2 
R 3v o 
R 
R
1 + 1  = v1 + 1 v 2
R3 + R 4  R 2 
R2
vo =
vO =

R3 + R 4 
R
 v1 + 1 v 2 
R2 

R 
R 3 1 + 1  
 R2 
R3 + R 4
( v1R 2 + v 2 )
R 3 ( R1 + R 2 )
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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COSMOS: Complete Online Solutions Manual Organization System
5.35
vo
R
= 1 + f = 10
vi
Ri
If Ri = 10kΩ, Rf = 90kΩ
Av =
Rf = 9Ri
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Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.36
VTh = Vab
But
R1
Vab . Thus,
R1 + R2
R
R + R2
= 1
v s = (1 + 2 )v s
R1
R1
vs =
VTh = Vab
To get RTh, apply a current source Io at terminals a-b as shown below.
v1
+
-
v2
a
+
R2
vo
io
R1
b
Since the noninverting terminal is connected to ground, v1 = v2 =0, i.e. no current passes
through R1 and consequently R2 . Thus, vo=0 and
v
RTh = o = 0
io
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.37
R

R
R
v o = −  f v1 + f v 2 + f v 3 
R2
R3 
 R1
30
30
 30

= −  (1) + (2) + (−3)
20
30
 10

vo = –3V
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5.38
R

R
R
R
v o = −  f v1 + f v 2 + f v 3 + f v 4 
R2
R3
R4 
 R1
50
50
50
 50

= −  (10) + (−20) + (50) + (−100)
20
10
50
 25

= -120mV
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.39
This is a summing amplifier.
Rf 
Rf
 Rf
50
50
 50

v3  = − (2) + v 2 + (−1)  = −9 − 2.5v 2
v2 +
v1 +
vo = −
R3 
20
50
R2
 10

 R1
Thus,
vo = −16.5 = −9 − 2.5v 2

→ v 2 = 3 V
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.40
Applying KCL at node a, where node a is the input to the op amp.
v1 − v a v 2 − v a v 3 − v a
+
+
= 0 or va = (v1 + v2 + v3)/3
R
R
R
vo = (1 + R1/R2)va = (1 + R1/R2)(v1 + v2 + v3)/3.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
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5.41
Rf/Ri = 1/(4)
Ri = 4Rf = 40kΩ
The averaging amplifier is as shown below:
v1
v2
v3
v4
R1 = 40 kΩ
10 kΩ
R2 = 40 kΩ
R3 = 40 kΩ
R4 = 40 kΩ
−
+
vo
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Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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COSMOS: Complete Online Solutions Manual Organization System
5.42
1
R f = R1 = 10 kΩ
3
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5-43
In order for

R
R
R
R
v o =  f v1 + f v 2 + f v 3 + f v 4 
R2
R3
R4 
 R1
to become
1
(v 1 + v 2 + v 3 + v 4 )
4
R
Rf 1
12
=
Rf = i =
= 3kΩ
Ri 4
4
4
vo = −
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.44
R4
R3
v1
v2
a
R1
−
+
b
R2
At node b,
v b − v1 v b − v 2
+
=0
R1
R2
At node a,
0 − va va − vo
=
R3
R4
vo
v1 v 2
+
R1 R 2
vb =
1
1
+
R1 R 2
(1)
vo
1+ R4 / R3
(2)
va =
But va = vb. We set (1) and (2) equal.
vo
R v + R 1v 2
= 2 1
1+ R4 / R3
R1 + R 2
or
vo =
(R 3 + R 4 )
(R 2 v1 + R 1v 2 )
R 3 (R 1 + R 2 )
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.45
This can be achieved as follows:
 R
(− v1 ) + R v 2 
v o = −
R/2 
R / 3
R

R
= −  f (− v1 ) + f v 2 
R2 
 R1
i.e. Rf = R, R1 = R/3, and R2 = R/2
Thus we need an inverter to invert v1, and a summer, as shown below (R<100kΩ).
R
v1
R
R
−
+
R/3
-v1
v2
R/2
−
+
vo
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.46
v1 1
R
R
R
1
+ ( − v 2 ) + v 3 = f v1 + x ( − v 2 ) + f v 3
3 3
2
R1
R2
R3
i.e. R3 = 2Rf, R1 = R2 = 3Rf. To get -v2, we need an inverter with Rf = Ri. If Rf = 10kΩ,
a solution is given below.
− vo =
10 kΩ
v2
v1
10 kΩ
−
+
30 kΩ
10 kΩ
30 kΩ
-v2
v3
20 kΩ
−
+
vo
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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COSMOS: Complete Online Solutions Manual Organization System
5.47
Using eq. (5.18), R1 = 2kΩ, R2 = 30kΩ, R3 = 2kΩ, R4 = 20kΩ
vo =
30(1+ 2 / 30)
30
32
v2 −
V1 =
(2) − 15(1) = 14.09 V
2(1+ 2 / 20)
2
2.2
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Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.48
We can break this problem up into parts. The 5 mV source separates the lower circuit
from the upper. In addition, there is no current flowing into the input of the op amp
which means we now have the 40-kohm resistor in series with a parallel combination of
the 60-kohm resistor and the equivalent 100-kohm resistor.
Thus,
40k + (60x100k)/(160) = 77.5k
which leads to the current flowing through this part of the circuit,
i = 5m/77.5k = 6.452x10–8
The voltage across the 60k and equivalent 100k is equal to,
v = ix37.5k = 2.419mV
We can now calculate the voltage across the 80-kohm resistor.
v80 = 0.8x2.419m = 1.9352mV
which is also the voltage at both inputs of the op amp aqnd the voltage between the 20kohm and 80-kohm resistors in the upper circuit. Let v1 be the voltage to the left of the
20-kohm resistor of the upper circuit and we can write a node equation at that node.
(v1–5m)/(10k) + v1/30k + (v1–1.9352m)/20k = 0
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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or
6v1 – 30m + 2v1 + 3v1 – 5.806m = 0
or
v1 = 35.806m/11 = 3.255mV
The current through the 20k-ohm resistor, left to right, is,
i20 = (3.255m–1.9352m)/20k = 6.599x10–8 A
thus,
vo = 1.9352m – 6.599x10–8x80k
= 1.9352m – 5.2792m = –3.344 mV.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.49
R1 = R3 = 10kΩ, R2/(R1) = 2
R2 = 2R1 = 20kΩ = R4
i.e.
Verify:
vo =
R 2 1 + R1 / R 2
R
v 2 − 2 v1
R1 1 + R 3 / R 4
R1
=2
(1 + 0.5)
v 2 − 2 v 1 = 2( v 2 − v 1 )
1 + 0.5
Thus, R1 = R3 = 10kΩ, R2 = R4 = 20kΩ
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.50
(a)
We use a difference amplifier, as shown below:
v1
v2
vo =
(b)
R1
R2
R1
−
+
R2
vo
R2
(v 2 − v1 ) = 2(v 2 − v1 ), i.e. R2/R1 = 2
R1
If R1 = 10 kΩ then R2 = 20kΩ
We may apply the idea in Prob. 5.35.
v 0 = 2 v1 − 2 v 2
 R
(− v1 ) + R v 2 
= −
R/2 
R / 2
R

R
= −  f (− v1 ) + f v 2 
R2 
 R1
i.e. Rf = R, R1 = R/2 = R2
We need an inverter to invert v1 and a summer, as shown below. We may let R = 10kΩ.
R
v1
R
R
−
+
R/2
-v1
v2
R/2
−
+
vo
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.51
We achieve this by cascading an inverting amplifier and two-input inverting summer as
shown below:
R
v1
R
R
−
+
R
va
v2
R
−
+
vo
Verify:
But
vo = -va - v2
va = -v1. Hence
vo = v1 - v2.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.52
A summing amplifier shown below will achieve the objective. An inverter is inserted to
invert v2. Let R = 10 k Ω .
R/2
R
v1
R/5
v3
v4
+
R
R
R
v2
-
R/4
+
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
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vo
COSMOS: Complete Online Solutions Manual Organization System
5.53
(a)
R1
v1
R2
va
vb
−
+
R2
R1
v2
vo
At node a,
At node b,
v1 − v a v a − v o
=
R1
R2
R2
vb =
v2
R1 + R 2
va =
R 2 v1 + R 1 v o
R1 + R 2
(1)
(2)
But va = vb. Setting (1) and (2) equal gives
R v + R 1v o
R2
v2 = 2 1
R1 + R 2
R1 + R 2
R
v 2 − v1 = 1 v o = v i
R2
vo R 2
=
vi
R1
(b)
−
v1
vi
+ v2
R1/2 v
A
R1/2
R2
va
Rg
R1/2
R1/2
vB
vb
−
+
R2
+
vo
−
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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At node A,
v1 − v A v B − v A v A − v a
+
=
R1 / 2
Rg
R1 / 2
or
v1 − v A +
At node B,
v2 − vB vB − vA vB − vb
=
+
R1 / 2
R1 / 2
Rg
or
v2 − vB −
R1
(v B − v A ) = v A − v a
2R g
R1
(v B − v A ) = v B − v b
2R g
(1)
(2)
Subtracting (1) from (2),
v 2 − v1 − v B + v A −
2R 1
(v B − v A ) = v B − v A − v b + v a
2R g
Since, va = vb,
v
R 
v 2 − v1 
= 1 + 1  (v B − v A ) = i
 2R 
2
2
g 

or
vB − vA =
vi
⋅
2
1
R
1+ 1
2R g
(3)
But for the difference amplifier,
R2
(v B − v A )
R1 / 2
R
vB − vA = 1 vo
2R 2
vo =
or
Equating (3) and (4),
v
R1
vo = i ⋅
2
2R 2
vo R 2
=
⋅
vi
R1
(4)
1
R
1+ 1
2R g
1
R
1+ 1
2R g
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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(c)
At node a,
At node b,
v1 − v a v a − v A
=
R1
R2 /2
2R 1
2R 1
v1 − v a =
va −
vA
R2
R2
2R 1
2R 1
v2 − vb =
vb −
vB
R2
R2
(1)
(2)
Since va = vb, we subtract (1) from (2),
v
− 2R 1
(v B − v A ) = i
R2
2
− R2
vB − vA =
vi
2R 1
v 2 − v1 =
or
(3)
At node A,
va − vA vB − vA vA − vo
+
=
R2 /2
Rg
R/2
va − vA +
At node B,
R2
(v B − v A ) = v A − v o
2R g
(4)
vb − vB vB − vA vB − 0
−
=
R/2
Rg
R/2
vb − vB −
R2
(v B − v A ) = v B
2R g
(5)
Subtracting (5) from (4),
v B −v A +
R2
(v B − v A ) = v A − v B − v o
Rg

R 
2(v B − v A )1 + 2  = − v o
 2R 
g 

Combining (3) and (6),
− R 2 
R 
v i 1 + 2  = −v o
 2R 
R1
g 

(6)
v o R 2 
R 
=
1+ 2 
vi
R 1  2R g 
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.54
The first stage is a summer (please note that we let the output of the first stage be v1).
R 
R
v1 = − v s + v o  = –vs – vo
R 
R
The second stage is a noninverting amplifier
vo = (1 + R/R)v1 = 2v1 = 2(–vs – vo) or 3vo = –2vs
vo/vs = –0.6667.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.55
Let A1 = k, A2 = k, and A3 = k/(4)
A = A1A2A3 = k3/(4)
20Log10 A = 42
Thus
Log10 A = 2.1
A = 102 ⋅1 = 125.89
k3 = 4A = 503.57
k = 3 503.57 = 7.956
A1 = A2 = 7.956, A3 = 1.989
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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COSMOS: Complete Online Solutions Manual Organization System
5.56
Each stage is an inverting amplifier. Hence.
vo
10
40
= (− )(− ) = 20
vs
1
20
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
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5.57
Let v1 be the output of the first op amp and v2 be the output of the second op amp.
The first stage is an inverting amplifier.
50
v1 = −
vs1 = −2vs1
25
The second state is a summer.
v2 = –(100/50)vs2 – (100/100)v1 = –2vs2 + 2vs1
The third state is a noninverting amplifier
100
)v2 = 3v2 = 6vs1 − 6vs2
vo = (1+
50
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.58
Looking at the circuit, the voltage at the right side of the 5-kΩ resistor must be at 0V if
the op amps are working correctly. Thus the 1-kΩ is in series with the parallel
combination of the 3-kΩ and the 5-kΩ. By voltage division, the input to the voltage
follower is:
v1 =
35
1+ 3 5
(0.6) = 0.3913V = to the output of the first op amp.
Thus
vo = –10((0.3913/5)+(0.3913/2)) = –2.739 V.
io =
0 − vo
= 0.6848 mA
4k
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.59
The first stage is a noninverting amplifier. If v1 is the output of the first op amp,
v1 = (1 + 2R/R)vs = 3vs
The second stage is an inverting amplifier
vo = –(4R/R)v1 = –4v1 = –4(3vs) = –12vs
vo/vs = –12.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
5.60
The first stage is a summer. Let V1 be the output of the first stage.
10
10
vi − vo

→ v1 = −2vi − 2.5vo
5
4
By voltage division,
10
5
v1 =
vo = vo
10 + 2
6
Combining (1) and (2),
5
10
vo = −2v1 − 2.5v0

→
v0 = −2vi
6
3
v1 = −
(1)
(2)
vo
= −6 /10 = −0.6
vi
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5.61
The first op amp is an inverter. If v1 is the output of the first op amp,
200
v1 = −
(0.4) = −0.8V
100
The second op amp is a summer
40
−40
(0.2) − (0.8) = 0.8 + 1.6 = 2.4 V
Vo =
10
20
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5.62
Let v1 = output of the first op amp
v2 = output of the second op amp
The first stage is a summer
v1 = −
R2
R
vi – 2 vo
R1
Rf
(1)
The second stage is a follower. By voltage division
vo = v2 =
R4
v1
R3 + R4
v1 =
R3 + R4
vo
R4
(2)
From (1) and (2),
 R3 
R
R
1 +
 v o = − 2 v i − 2 v o
Rf
R1
 R4 
 R3 R2 
R
 v o = − 2 v i
1 +
+
R1
 R4 Rf 
vo
R
− R 2R 4R f
1
=
=− 2 ⋅
R
R
R 1 (R 2 R 4 + R 3 R f + R 4 R f )
vi
R1
1+ 3 + 2
R4 Rf
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5.63
The two op amps are summers. Let v1 be the output of the first op amp. For the first
stage,
v1 = −
R2
R
vi − 2 vo
R1
R3
(1)
For the second stage,
vo = −
R4
R
v1 − 4 v i
R5
R6
(2)
Combining (1) and (2),
 R2 
R R 
R

 v i + 4  2  v o − 4 v i
R5  R3 
R6
 R1 
 R R  R R
R 
v o 1 − 2 4  =  2 4 − 4  v i
 R 3 R 5   R 1R 5 R 6 
vo =
R4
R5
R 2R 4 R 4
−
vo
R 1R 5 R 6
=
R R
vi
1− 2 4
R 3R 5
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5.64
G4
G
G3
G1
1
+
0V
G
+
vs
v
2
0V +
G2
+
vo
-
At node 1, v1=0 so that KCL gives
G1v s + G4 vo = −Gv
(1)
At node 2,
G2 v s + G3 v o = −Gv
From (1) and (2),
G1v s + G4 v o = G2 v s + G3 vo
or
vo G1 − G2
=
v s G3 − G 4
(2)
→

(G1 − G2 )v s = (G3 − G4 )vo
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5.65
The output of the first op amp (to the left) is 6 mV. The second op amp is an inverter so
that its output is
30
(6mV) = -18 mV
10
The third op amp is a noninverter so that
vo ' = −
vo ' =
40
vo
40 + 8

→
vo =
48
vo ' = − 21.6 mV
40
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5.66
100  40 
− 100
100
(6) −
(2)
 − (4) −
25
20  20 
10
= −24 + 40 − 20 = -4V
vo =
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5.67
80  80 
80
 − (0.2) − (0.2)
40  20 
20
= 3.2 − 0.8 = 2.4V
vo = −
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5.68
If Rq = ∞, the first stage is an inverter.
Va = −
15
(10) = −30mV
5
when Va is the output of the first op amp.
The second stage is a noninverting amplifier.
 6
v o = 1 +  v a = (1 + 3)(−30) = -120mV
 2
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5.69
In this case, the first stage is a summer
va = −
15
15
(10) − v o = −30 − 1.5v o
5
10
For the second stage,
 6
v o = 1 +  v a = 4 v a = 4(− 30 − 1.5v o )
 2
120
7 v o = −120
vo = −
= -17.143mV
7
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5.70
The output of amplifier A is
vA = −
30
30
(1) − (2) = −9
10
10
The output of amplifier B is
vB = −
20
20
(3) − (4) = −14
10
10
40 kΩ
vA
vB
20 kΩ
a
60 kΩ
−
+
b
vo
10 kΩ
vb =
10
(−14) = −2V
60 + 10
At node a,
vA − va va − vo
=
20
40
But va = vb = -2V, 2(-9+2) = -2-vo
Therefore,
vo = 12V
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5.71
20k Ω
5k Ω
100k Ω
40k Ω
+
+
2V
-
v2
10k Ω
80k Ω
+
20k Ω
+
vo
-
+
+
3V
-
10k Ω
v1
+
-
30k Ω
v3
50k Ω
20
50
(2) = −8, v3 = (1 + )v1 = 8
5
30
100 
 100
vo = −
v2 +
v3  = −(−20 + 10) = 10 V
80 
 40
v1 = 3,
v2 = −
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5.72
Since no current flows into the input terminals of ideal op amp, there is no voltage drop
across the 20 kΩ resistor. As a voltage summer, the output of the first op amp is
v01 = 0.4
The second stage is an inverter
250
v 01
100
= −2.5(0.4) = -1V
v2 = −
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5.73
The first stage is an inverter. The output is
50
v 01 = − (−1.8) + 1.8 = 10.8V
10
The second stage is
v 2 = v 01 = 10.8V
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5.74
Let v1 = output of the first op amp
v2 = input of the second op amp.
The two sub-circuits are inverting amplifiers
100
(0.6) = −6V
10
32
v2 = −
(0.4) = −8V
1.6
v − v2
−6+8
io = 1
=−
= 100 µA
20k
20k
v1 = −
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5.75
The schematic is shown below. Pseudo-components VIEWPOINT and IPROBE are
involved as shown to measure vo and i respectively. Once the circuit is saved, we click
Analysis | Simulate. The values of v and i are displayed on the pseudo-components as:
i = 200 µA
(vo/vs) = -4/2 = –2
The results are slightly different than those obtained in Example 5.11.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
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5.76
The schematic is shown below. IPROBE is inserted to measure io. Upon simulation, the
value of io is displayed on IPROBE as
io = -374.78 µA
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5.77
The schematic for the PSpice solution is shown below.
Note that the output voltage, –3.343 mV, agrees with the answer to problem, 5.48.
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5.78
The circuit is constructed as shown below. We insert a VIEWPOINT to display vo.
Upon simulating the circuit, we obtain,
vo = 667.75 mV
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5.79
The schematic is shown below. A pseudo-component VIEWPOINT is inserted to display
vo. After saving and simulating the circuit, we obtain,
vo = -14.61 V
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5.80
The schematic is as shown below. After it is saved and simulated, we obtain
vo = 2.4 V.
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5.81
The schematic is shown below. We insert one VIEWPOINT and one IPROBE to
measure vo and io respectively. Upon saving and simulating the circuit, we obtain,
vo = 343.4 mV
io = 24.51 µA
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5.82
The maximum voltage level corresponds to
11111 = 25 – 1 = 31
Hence, each bit is worth
(7.75/31) = 250 mV
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5.83
The result depends on your design. Hence, let RG = 10 k ohms, R1 = 10 k ohms, R2 =
20 k ohms, R3 = 40 k ohms, R4 = 80 k ohms, R5 = 160 k ohms, R6 = 320 k ohms,
then,
-vo = (Rf/R1)v1 + --------- + (Rf/R6)v6
= v1 + 0.5v2 + 0.25v3 + 0.125v4 + 0.0625v5 + 0.03125v6
(a)
|vo| = 1.1875 = 1 + 0.125 + 0.0625 = 1 + (1/8) + (1/16) which implies,
[v1 v2 v3 v4 v5 v6] = [100110]
(b)
|vo| = 0 + (1/2) + (1/4) + 0 + (1/16) + (1/32) = (27/32) = 843.75 mV
(c)
This corresponds to [1 1 1 1 1 1].
|vo| = 1 + (1/2) + (1/4) + (1/8) + (1/16) + (1/32) = 63/32 = 1.96875 V
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5.84
For (a), the process of the proof is time consuming and the results are only approximate,
but close enough for the applications where this device is used.
(a)
The easiest way to solve this problem is to use superposition and to solve
for each term letting all of the corresponding voltages be equal to zero.
Also, starting with each current contribution (ik) equal to one amp and
working backwards is easiest.
2R
v1
+
−
R
R
R
2R
ik
v2
+
−
2R
v3
+
−
2R
v4
+
−
R
For the first case, let v2 = v3 = v4 = 0, and i1 = 1A.
Therefore,
v1 = 2R volts or i1 = v1/(2R).
Second case, let v1 = v3 = v4 = 0, and i2 = 1A.
Therefore,
v2 = 85R/21 volts or i2 = 21v2/(85R). Clearly this is not
th
(1/4 ), so where is the difference? (21/85) = 0.247 which is a really
good approximation for 0.25. Since this is a practical electronic circuit,
the result is good enough for all practical purposes.
Now for the third case, let v1 = v2 = v4 = 0, and i3 = 1A.
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Therefore,
v3 = 8.5R volts or i3 = v3/(8.5R). Clearly this is not
th
(1/8 ), so where is the difference? (1/8.5) = 0.11765 which is a really
good approximation for 0.125. Since this is a practical electronic circuit,
the result is good enough for all practical purposes.
Finally, for the fourth case, let v1 = v2 = v4 = 0, and i3 = 1A.
Therefore,
v4 = 16.25R volts or i4 = v4/(16.25R). Clearly this is not
(1/16th), so where is the difference? (1/16.25) = 0.06154 which is a
really good approximation for 0.0625. Since this is a practical electronic
circuit, the result is good enough for all practical purposes.
Please note that a goal of a lot of electronic design is to come up with
practical circuits that are economical to design and build yet give the
desired results.
(a)
If Rf = 12 k ohms and R = 10 k ohms,
-vo = (12/20)[v1 + (v2/2) + (v3/4) + (v4/8)]
= 0.6[v1 + 0.5v2 + 0.25v3 + 0.125v4]
For
[v1 v2 v3 v4] = [1 0 11],
|vo| = 0.6[1 + 0.25 + 0.125] = 825 mV
For
[v1 v2 v3 v4] = [0 1 0 1],
|vo| = 0.6[0.5 + 0.125] = 375 mV
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5.85
This is a noninverting amplifier.
vo = (1 + R/40k)vs = (1 + R/40k)2
The power being delivered to the 10-kΩ give us
P = 10 mW = (vo)2/10k or vo = 10 − 2 x10 4 = 10V
Returning to our first equation we get
10 = (1 + R/40k)2 or R/40k = 5 – 1 = 4
Thus,
R = 160 kΩ.
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5.86
vo = A(v2 – v1) = 200(v2 – v1)
(a)
vo = 200(0.386 – 0.402) = -3.2 V
(b)
vo = 200(1.011 – 1.002) = 1.8 V
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5.87
The output, va, of the first op amp is,
Also,
va = (1 + (R2/R1))v1
(1)
vo = (-R4/R3)va + (1 + (R4/R3))v2
(2)
Substituting (1) into (2),
vo = (-R4/R3) (1 + (R2/R1))v1 + (1 + (R4/R3))v2
Or,
If
vo = (1 + (R4/R3))v2 – (R4/R3 + (R2R4/R1R3))v1
R4 = R1 and R3 = R2, then,
vo = (1 + (R4/R3))(v2 – v1)
which is a subtractor with a gain of (1 + (R4/R3)).
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5.88
We need to find VTh at terminals a – b, from this,
vo = (R2/R1)(1 + 2(R3/R4))VTh = (500/25)(1 + 2(10/2))VTh
= 220VTh
Now we use Fig. (b) to find VTh in terms of vi.
a
a
30 kΩ
20 kΩ
20 kΩ
vi
vi
30 kΩ
+−
40 kΩ
80 kΩ
40 kΩ
80 kΩ
b
b
(b)
(a)
va = (3/5)vi, vb = (2/3)vi
VTh = vb – va (1/15)vi
(vo/vi) = Av = -220/15 = -14.667
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5.89
A summer with vo = –v1 – (5/3)v2 where v2 = 6-V battery and an inverting amplifier
with v1 = –12vs.
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5.90
Transforming the current source to a voltage source produces the circuit below,
vb = (2/(2 + 4))vo = vo/3
At node b,
20 kΩ
5 kΩ a
5is
+
−
b
−
+
4 kΩ
io
2 kΩ
+
vo
−
At node a,
(5is – va)/5 = (va – vo)/20
But va = vb = vo/3.
20is – (4/3)vo = (1/3)vo – vo, or is = vo/30
io = [(2/(2 + 4))/2]vo = vo/6
io/is = (vo/6)/(vo/30) = 5
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5.91
−
+
vo
R2
R1
is
i2
i1
But
io
io = i1 + i2
(1)
i1 = is
(2)
R1 and R2 have the same voltage, vo, across them.
R1i1 = R2i2, which leads to i2 = (R1/R2)i1
(3)
Substituting (2) and (3) into (1) gives,
io = is(1 + R1/R2)
io/is = 1 + (R1/R2) = 1 + 8/1 = 9
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5.92
The top op amp circuit is a non-inverter, while the lower one is an inverter. The output
at the top op amp is
v1 = (1 + 60/30)vi = 3vi
while the output of the lower op amp is
v2 = -(50/20)vi = -2.5vi
Hence,
vo = v1 – v2 = 3vi + 2.5vi = 5.5vi
vo/vi = 5.5
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5.93
R3
R1 v
a
vb
−
+
io
R4
+
vi
+
−
R2
vL
iL
RL
+
vo
−
−
At node a,
(vi – va)/R1 = (va – vo)/R3
vi – va = (R1/R2)(va – vo)
vi + (R1/R3)vo = (1 + R1/R3)va
(1)
But va = vb = vL. Hence, (1) becomes
vi = (1 + R1/R3)vL – (R1/R3)vo
(2)
io = vo/(R4 + R2||RL), iL = (R2/(R2 + RL))io = (R2/(R2 + RL))(vo/( R4 + R2||RL))
Or,
vo = iL[(R2 + RL)( R4 + R2||RL)/R2
(3)
But,
vL = iLRL
(4)
Substituting (3) and (4) into (2),
vi = (1 + R1/R3) iLRL – R1[(R2 + RL)/(R2R3)]( R4 + R2||RL)iL
= [((R3 + R1)/R3)RL – R1((R2 + RL)/(R2R3)(R4 + (R2RL/(R2 + RL))]iL
= (1/A)iL
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
COSMOS: Complete Online Solutions Manual Organization System
Thus,
A =
1

 R + RL
R 
 1 + 1  R L − R 1  2
R3 

 R 2R 3

R 2RL
 R 4 +
R2 + RL




Please note that A has the units of mhos. An easy check is to let every resistor equal 1ohm and vi equal to one amp. Going through the circuit produces iL = 1A. Plugging into
the above equation produces the same answer so the answer does check.
C.K. Alexander, M.N.O. Sadiku, Circuiti elettrici, 4e © 2014 McGraw-Hill Education (Italy) srl
Fundamentals of Electric Circuits, 3/e, Charles Alexander, Matthew Sadiku
© 2007 The McGraw-Hill Companies.